The system of linear equations in real (x, y) given by $\left(\begin{array}{cc}\mathrm{x}& \mathrm{y}\end{array}\right)\left[\begin{array}{cc}2& 5-2\alpha \\ \alpha & 1\end{array}\right]=\left(\begin{array}{cc}0& 0\end{array}\right)$ involves a real parameter α and has infinitely many non-trivial solutions for special value(s) of α. Which one or more among the following options is/are non-trivial solution(s) of (x, y) for such special value(s) of α ?

The correct options are:
x = 2, y = – 2 and x = –1, y = 4.

Let's analyze the given system of linear equations step-by-step to understand why these options are the correct non-trivial solutions.

The given matrix equation is:
$\left(\begin{array}{cc}x& y\end{array}\right)\left[\begin{array}{cc}2& 5-2\alpha \\ \alpha & 1\end{array}\right]=\left(\begin{array}{cc}0& 0\end{array}\right)$

By performing the matrix multiplication on the left-hand side, we obtain the following system of linear equations:
1) $2x+\alpha y=0$
2) $(5-2\alpha )x+y=0$

This is a homogeneous system of linear equations in variables $x$ and $y$. For a homogeneous system of equations to have infinitely many non-trivial solutions, the determinant of the coefficient matrix (transpose or direct, which has the same determinant) must be zero. Alternatively, we can write the system as:
$\left[\begin{array}{cc}2& \alpha \\ 5-2\alpha & 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Setting the determinant of the coefficient matrix to zero:
$\left|\begin{array}{cc}2& \alpha \\ 5-2\alpha & 1\end{array}\right|=0$

Expanding the determinant:
$2-\alpha (5-2\alpha )=0$
$2-5\alpha +2{\alpha }^2=0$
$2{\alpha }^2-5\alpha +2=0$

Factoring the quadratic equation:
$2{\alpha }^2-4\alpha -\alpha +2=0$
$2\alpha (\alpha -2)-1(\alpha -2)=0$
$(2\alpha -1)(\alpha -2)=0$

This gives two special values of $\alpha$:
Case 1: $\alpha =2$
Case 2: $\alpha =\frac{1}{2}$

Now we analyze the solutions $(x,y)$ corresponding to these cases:

Case 1: For $\alpha =2$
Substituting $\alpha =2$ into Equation 1:
$2x+2y=0\Rightarrow y=-x$
Any solution must satisfy $y=-x$. Let's check the given options:

• For x = 2, y = – 2: This satisfies $y=-x$ (since $-2=-\left(2\right)$). Thus, this is a non-trivial solution.

Case 2: For $\alpha =\frac{1}{2}$
Substituting $\alpha =\frac{1}{2}$ into Equation 1:
$2x+\frac{1}{2}y=0\Rightarrow y=-4x$
Any solution must satisfy $y=-4x$. Let's check the given options:

• For x = –1, y = 4: This satisfies $y=-4x$ (since $4=-4(-1)$). Thus, this is also a non-trivial solution.

Therefore, the options x = 2, y = – 2 and x = –1, y = 4 represent valid non-trivial solutions for the system under the special values of $\alpha$.